Hyperbolic area and $SL_2$ Given that $\mu(A) := \iint_{A}\frac{\mathrm dx\mathrm dy}{y^2}$ where $A \subset H$ and $H$ is the upper half-plane, I need to show that:
a. The measure $\mu$ is invariant under all $g \in SL_2(\mathbb R)$ i.e. $\mu(gA) = \mu(A)$.
b. Compute the hyperbolic area of the standard fundamental domain for the modular group $SL_2(\mathbb Z)$ ( $ A = \{\tau | -0.5 \leq \Re(\tau) \leq 0.5 , |\tau| \geq 1\} $ ).
 A: Hints:


*

*That the measure is (left) invariant is essentially shown here.

*You want to compute the integral $$\mu(A) = \int_{-\frac{1}{2}}^{\frac{1}{2}} \int_{\sqrt{1-x^2}}^{\infty} \frac{1}{y^2}\,dy\,dx$$
and that shouldn't be too hard. To check your computations, the answer is $\frac{\pi}{3}$.

On Pavel's request I elaborate on point 1.
Write $z = x + iy$ and $w = g\cdot z = \frac{az + b}{cz + d} = u + iv$ where $g = \begin{bmatrix}a & b \\\ c & d \end{bmatrix}\in \operatorname{SL}(2,\mathbb{R})$. Then the Jacobian determinant is
$$\frac{\partial (u,v)}{\partial(x,y)} =
\frac{\partial u}{\partial x} \frac{\partial v}{\partial y} -
\frac{\partial u}{\partial y} \frac{\partial v}{\partial x}$$
and by appealing to the Cauchy-Riemann equations we see that this is equal to 
$$\frac{\partial (u,v)}{\partial(x,y)} =
\left(\frac{\partial u}{\partial x}\right)^2 +
\left(\frac{\partial v}{\partial x}\right)^2 = \cdots = \frac{1}{|cz + d|^4}$$
using the calculations in the Wikipedia link I gave above.
Recalling that $v = \frac{y}{|cz + d|^2}$ we can compute
$$
\mu(gA) = \int\!\!\!\!\int_{gA} \frac{du\,dv}{v^2} =
\int\!\!\!\!\int_{A}
\frac{|cz + d|^4}{y^2}\,\frac{\partial (u,v)}{\partial(x,y)}\, dx\, dy = \int\!\!\!\!\int_{A} \frac{dx\, dy}{y^2} = \mu(A)$$
as we wanted.
